\(\sqrt{3}< \sqrt{4}=2\)

\(5-\sqrt{8}=5-2\sqrt{2}\)

Mà :

\(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)

Nên :

\(5-2\sqrt{2}>5-3=2\)

Vậy :

\(\sqrt{3}< 5-\sqrt{8}\)

√3<√4=23<4=2

5−√8=5−2√25−8=5−22 mà

2√2=√8<√9=322=8<9=3nFe=nFe2On=a(mol)nên 5−2√2>5−3=25−22>5−3=256a+a(112+16n)=14,4(1)

Vậy √3<5−√8nSO2=0,1(mol)

k cho mk nhé, cảm ơn

đk: \(x\ne-1\)

\(PT\Leftrightarrow x^2-\frac{2x^2}{x+1}+\left(\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}=1\)

\(\Leftrightarrow\left(x-\frac{x}{x+1}\right)^2+2\frac{x^2}{x+1}=1\)

\(\Leftrightarrow\left(\frac{x^2}{x+1}+1\right)^2=2\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+\left(1-\sqrt{2}\right)x+\left(1-\sqrt{2}\right)=0\\x^2+\left(1+\sqrt{2}\right)x+\left(1+\sqrt{2}\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{\sqrt{2}-1+\sqrt{2\sqrt{2}-1}}{2}\\\frac{\sqrt{2}-1-\sqrt{2\sqrt{2}-1}}{2}\end{cases}\left(TM\right)}\)

Theo hđt : \(\left(a-b\right)^2=a^2-2ab+b^2\Rightarrow a^2+b^2=\left(a-b\right)^2+2ab\)

pt có dạng : \(\left(x-\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}=1\)ĐK : \(x\ne1\)

\(\Leftrightarrow\left(\frac{x^2+x-x}{x+1}\right)^2+\frac{2x^2}{x+1}=1\Leftrightarrow\frac{x^4+2x^2}{x+1}=\frac{x+1}{x+1}\)

\(\Rightarrow x^4+2x^2-x-1=0\Rightarrow x_1=-0,48...;x_2=0,82...\)( tmđk ) 

\(\frac{-19\sqrt{6-46}}{5}\approx-18.508061\) 

Chúc anh học tốt!

địt ko em

a. \(\sqrt{4x}+\sqrt{x}=2\Leftrightarrow2\sqrt{x}+\sqrt{x}=2\Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\frac{2}{3}\Leftrightarrow x=\frac{4}{9}\)

b. \(\sqrt{x^2-4}=\sqrt{x-2}\Leftrightarrow\hept{\begin{cases}x^2-4=x-2\\x-2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}\orbr{\begin{cases}x=2\\x=-1\end{cases}}\\x\ge2\end{cases}}\Leftrightarrow x=2\)\(\sqrt{x^2-4}=\sqrt{x-2}\Leftrightarrow\hept{\begin{cases}x^2-4=x-2\\x-2\ge2\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x-2\right)\left(x+1\right)=0\\x\ge2\end{cases}}\Leftrightarrow x=2\)

c.\(\sqrt{x^2-2x}+\sqrt{2x^2+4x}=2x\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2-2x+2x^2+4x+2\sqrt{x^2-2x}.\sqrt{2x^2+4x}=4x^2\end{cases}}\)

\(\Rightarrow x^2-2x=2\sqrt{x^2-2x}.\sqrt{2x^2+4x}\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-2x}=0\\\sqrt{x^2-2x}=2\sqrt{2x^2+4x}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\text{ hoặc }x=2\\x^2-2x=8x^2+16x\end{cases}\Leftrightarrow}\)hoặc x=0 hoặc x=2 hoặc x= -18/7

Kết hợp điều kiện ta có : \(x=0\text{ hoặc }x=2\)

d. Điều kiện \(x\ge3\) ta có :

\(\sqrt{x^2+2x-15}=\sqrt{x-3}+\sqrt{x^2-3x}\Leftrightarrow x^2+2x-15=x^2-2x-3+2\sqrt{x-3}\sqrt{x^2-3x}\)

\(\Leftrightarrow2x-6=\sqrt{x-3}.\sqrt{x^2-3x}\Leftrightarrow4\left(x-3\right)^2=\left(x-3\right)\left(x^2-3x\right)\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)

a,b,c>0

\(\frac{2}{4\sqrt{3}-7}+\frac{2}{4\sqrt{3}+7}\)

\(=-14-8\sqrt{3}+14-8\sqrt{3}\)

\(=-16\sqrt{3}\)

ta có \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3\)

ta có :\(\sqrt{4x+5}=\frac{\sqrt{9\left(4x+5\right)}}{3}\le\frac{9+4x+5}{2\times3}=\frac{2x+7}{3}\)

tương tự ta sẽ có  ; \(A\le\frac{2x+7}{3}+\frac{2y+7}{3}+\frac{2z+7}{3}=\frac{2}{3}\left(x+y+z\right)+7\le\frac{2}{3}\times3+7=9\)

Vậy GTLN của A=9

dấu bằng xảy ra khi x= y= z =1

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=3.3=9\)

\(\Rightarrow x+y+z\le3\).

\(A=\sqrt{4x+5}+\sqrt{4y+5}+\sqrt{4z+5}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4x+5+4y+5+4z+5\right)}\)

\(=\sqrt{3\left[4\left(x+y+z\right)+15\right]}=9\)

Dấu \(=\)khi \(x=y=z=1\).